Question

An observer standing 50 m away from a building notices that the angles of elevation of the top and bottom of a flagstaff on the building are 60° and 45° respectively. The height of the flagstaff is

(a) $50\sqrt{3}\mathrm{m}$
(b) $50(\sqrt{3}+1)\mathrm{m}$
(c) $50(\sqrt{3}-1)\mathrm{m}$
(d) $\frac{50}{\sqrt{3}}\mathrm{m}$

Answer 1

(c) $50(\sqrt{3}-1)\mathrm{m}$
Let $AB$ be the tower, $BC$ be the flagstaff and O be the position of the observer. Thus, we have:
$OA=50$ m, ∠$AOB={45}^o$ and ∠$AOC={60}^o$

Now, in ∆$AOB$, we have:
$\frac{AB}{OA}=\tan {45}^o=1$

⇒ $\frac{AB}{50}=1$
⇒ $AB=50$ m
In ∆$AOC$, we have:
$\frac{AC}{OA}=\tan {60}^o=\sqrt{3}$

⇒ $\frac{AC}{50}=\sqrt{3}$
⇒ $AC=50\sqrt{3}$ m

∴ Height of the flagstaff = $BC=(AC-AB)=(50\sqrt{3}-50)=50(\sqrt{3}-1)$ m